1)Let *A=\(\large(a^{2}+4ab+b^{2})\)*, where *a* and *b* are postive integers.

Prove that **A≠2015**. **(HINT-Use Modular Arithmetic or Try to use the concept of Odd-Even numbers)**

2)Prove that any prime number, **\(\large(2^{2^n}+1)\)** cannot be represented as a difference of 2 fifth powers of integers.**(HINT-Expand the expression and You will get something in common)**

3)Find all pairs *(x,y)* where *x* and *y* are integers such that *\(\large(x^{3}+11^{3}=y^{3})\)*.**(HINT-Try to apply identities)**

4)If *\(\large(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c})\)*

Then Prove that,

*\(\large(\frac{1}{a^{3}}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^{3}+b^{3}+c^{3}})\)*.**(HINT-Factorize )**

5)Prove that *\(\large(2222^{5555}+5555^{2222})\)* is divisible by 7.**(HINT-Use Modular Arithmetic or Apply the concept of \(a^{n}±b^{n}\).)**

Today I have posted some easy problems ,So I'm sure that you will be able to solve all these 5 problems in 1.5 hours.

Also try my Set RMO.

## Comments

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TopNewestHere's my solution to

(1).Assume that ,

\(\large (a+b)^{2}+2ab=2015\)\(\large (a+b)^{2}=2015-2ab\)So,

\(\large (a+b)^{2}\)=OddSo,

\(\large (a+b)\)=Odd(2q+1)For some positive integer q.So,

\(\large 2ab=4p\)For some integer p.So,

\(\large (2q+1)^{2}+4p=2015\)\(\large 4(q^{2}+q+p)=2014\)\(\large q^{2}+q+p=\frac{2014}{4}\)Now we can see that R.H.S.is not an integer but In L.H.S. all are intgers .Thus it does not satisfy the given conditions.Hence Proved. – Naitik Sanghavi · 1 year, 8 months ago

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4) let \(p(x) = x^3 + lx^2 + mx + p\) with roots a,b,c then \(p = mn\) and \(c = -b\) . Now it easy – Dev Sharma · 1 year, 8 months ago

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– Naitik Sanghavi · 1 year, 8 months ago

Nice!!You got a different to solve this!Log in to reply

5) \(5555=4 \pmod 7\) or \(5555^{2222}=4^{2222} \pmod 7\)

\(4^3=1 \pmod 7\) or \(4^{2220}=1 \pmod 7\) or \(4^{2222}=16=2 \pmod 7\)

Therefore, \(5555^{2222}=2 \pmod 7\).

Similarly,

\(2222=3 \pmod 7\) or \(2222^{5555}=3^{5555} \pmod 7\)

\(3^3=-1 \pmod 7\) or \(3^{5553}=(-1)^{1851}=-1=6 \pmod 7\) or \(3^{5555}=54=5 \pmod 7\)

Therefore, \(5555^{2222}=5 \pmod 7\)

Therefore \(5555^{2222}+2222^{5555}=2+5=7=0 \pmod 7\) – Svatejas Shivakumar · 1 year, 8 months ago

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– Naitik Sanghavi · 1 year, 8 months ago

It seems that you have solved all problems of this note,You can my other problems of this set.Try my set Olympiad Preparation Set for an easier level !!Log in to reply

2) Suppose \(2^{2^n}+1=a^5-b^5\),where \(a^5-b^5\) be the difference of two fifth powers.\(a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)\ \quad where \quad a>b\).

Since \(2^{2^n}+1 \quad is \quad prime, \quad a-b=1\).

By Fermat's Little theorem, \(2^{2^n}+1=a^5-b^5=a-b \pmod 5=1 \pmod 5\).

Thus \(2^{2^n}+1\) cannot be represented as a difference of two fifth powers.

Moreover, \(2^{2^n}+1\) cannot be represented as the difference of two pth powers, where p is an odd prime. – Svatejas Shivakumar · 1 year, 8 months ago

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1) \(a^2 + 4ab + b^2 = 0, 2 or 1 mod4\) but 2015 = 3 mod4. A contradiction. – Dev Sharma · 1 year, 8 months ago

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– Naitik Sanghavi · 1 year, 8 months ago

I seems that you have solved all problems on this Set,You can try my other problems of this set and Try my set Olympiad Preparation setfor an easier level!!Log in to reply

– Svatejas Shivakumar · 1 year, 8 months ago

Just a small correction, \(a^2+4ab+b^2\) can also be \(0 \pmod 4\).Try using \pmod for mod. I too did it exactly the same way.Log in to reply

– Naitik Sanghavi · 1 year, 8 months ago

Yes,of course!! You can post your solutions if you have solved any or have an different solution from others!!!Log in to reply

– Naitik Sanghavi · 1 year, 8 months ago

Nice Observation!! I'm having an different solution!!You can try out other problems!!I will post my solutions after some days!!!!!!!Log in to reply

3) Is (-11,0),(0,11) the only solutions? – Svatejas Shivakumar · 1 year, 8 months ago

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– Naitik Sanghavi · 1 year, 8 months ago

Yes!!Correct but you need to prove that there are no other solutions too!!Log in to reply

@naitik sanghavi \(y= \sqrt [3]{x^3+11^3}=\sqrt [3]{(x+11)(x^2-11x+121)}\) which is only a perfect cube when either of the terms are zero. Hence, getting the above solution. – Svatejas Shivakumar · 1 year, 8 months ago

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4) This one is quite long! After a lot of solving I got c=-a or c=-b . Substitute both the values and you get the second equation. Am I missing an easier method? – Svatejas Shivakumar · 1 year, 8 months ago

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\(\large \frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)\(\large \frac{a+b}{ab}=\frac{c-(a+b+c)}{(a+b+c)×c}\)\(\large (a+b)×c×(a+b+c)=-(a+b)×ab\)(Don't remove (a+b) from both sides!Why?See Ahead!!!)\(\large (a+b)×c×(a+b+c)+(a+b)ab=0\)\(\large (a+b)[c×(a+c)+b×(a+c)]=0\)\(\large (a+b)(b+c)(c+a)=0\)a=-b or b=-c or c=-a.Now substitute anyone value and it will proved!! – Naitik Sanghavi · 1 year, 8 months ago

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@naitik sanghavi Nice!!.I multiplied (ab+bc+ca) with (a+b+c),did a little bit of factoring and got \(c^2+(a+b)c+ab=0\) and finally solved the quadratic equation to get the above solution.Your solution is much faster upvoted!! – Svatejas Shivakumar · 1 year, 8 months ago

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– Priyanshu Mishra · 1 year, 7 months ago

You can also generalize it for powers of n and then by induction it will be proved for 3.Log in to reply

Thanks for your contribution for the Brilliant RMO community :) – Swapnil Das · 1 year, 8 months ago

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– Naitik Sanghavi · 1 year, 8 months ago

Its OK .I am also preparing for RMO so just posting some problems daily..!!!!Log in to reply